Question: Evaluate the double integral. $ \int_1^3 \int_{-\sqrt{y}}^{\sqrt{y}} x^3 + 2 \, dx \, dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $4\sqrt{3} - \dfrac{4}{3}$ (Choice B) B $2\sqrt{3} - \dfrac{2}{3}$ (Choice C) C $8\sqrt{3} - \dfrac{8}{3}$ (Choice D) D $\sqrt{3} - \dfrac{1}{3}$
Answer: First, we evaluate the inner integral. We can substitute in the $-\sqrt{y}$ and $\sqrt{y}$ at the end as if they were numerical bounds. $\begin{aligned} \int_1^3 \int_{-\sqrt{y}}^{\sqrt{y}} x^3 + 2 \, dx \, dy &= \int_1^3 \left[ \dfrac{x^4}{4} + 2x \right]_{-\sqrt{y}}^{\sqrt{y}} dy \\ \\ &= \int_1^3 4\sqrt{y} \, dy \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_1^3 4\sqrt{y} \, dy &= 4 \left[ \dfrac{y^{1.5}}{1.5} \right]_1^3 \\ \\ &= 4 \left( \dfrac{6 \sqrt{3}}{3} - \dfrac{2}{3} \right) \\ \\ &= 8\sqrt{3} - \dfrac{8}{3} \end{aligned}$ The answer: $ \int_1^3 \int_{-\sqrt{y}}^{\sqrt{y}} x^3 + 2 \, dx \, dy = 8\sqrt{3} - \dfrac{8}{3}$